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2 years ago

9

A sample initially contains 8.0 moles of a radioactive isotope. How much of the sample remains after four half-lives? Express yo

ur answer using two significant figures.

1 answer:

Tju [1.3M]2 years ago

6 0

Answer:

0.50 mol

Explanation:

The half-life is <em>the time required for the amount of a radioactive isotope to decay to half that amount</em>.

Initially, there are 8.0 moles.

After 1 half-life, there remain 1/2 × 8.0 mol = 4.0 mol.

After 2 half-lives, there remain 1/2 × 4.0 mol = 2.0 mol.

After 3 half-lives, there remain 1/2 × 2.0 mol = 1.0 mol.

After 4 half-lives, there remain 1/2 × 1.0 mol = 0.50 mol.

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C) Mercury is solid at room temperature....True Or False<br>

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Answer:

true

Explanation:

Because Mercury can be solidified when its temparature us brought to its freezing point. However, when returned to room temparature conditions, mercury does not exist in solid state for long, and returns back to its more common liquid form.

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3 years ago

Filtration definition

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3 years ago

Read 2 more answers

Student have a receipt which calls for 1.15 lb of flour, but, of course, in France you will need to purchase flour in kilograms.

Verizon [17]

0.52164 kg of Flour

<u>Explanation:</u>

We have to convert lb (pounds) kg (kilograms), by using the conversion unit,

1 lb = 0.4536 kg

So to convert 1.15 lb into kg, we have to multiply 1.15 by 0.4536.

$1.15 \times 0.4536$ = 0.52164 kg of Flour

So the receipt calls for 0.52164 kg of Flour.

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3 years ago

You are going to standardize your sodium hydroxide by titrating with potassium hydrogen phthalate. As an example, you dissolve 0

Kay [80]

Answer:

0.13 M

Explanation:

The reaction equation is;

NaOH(aq) + KHC8H4O4(aq) ------> KNaC8H4O4(aq) + H2O(l)

Molar mass of KHP = 204.22 g/mol

Amount of KHP= mass/ molar mass = 0.3365 g/204.22 g/mol = 1.65 × 10^-3 moles

n= CV

Where;

C= concentration

V= volume in dm^3

n= number of moles

C= n/V = 1.65 × 10^-3 moles × 1000/250 = 6.6 × 10^-3 M

If 1 mole of KHP reacts with 1 mole of NaOH

1.65 × 10^-3 moles of KHP will react with 1.65 × 10^-3 moles of NaOH

From

n= CV

We have that only 12.44 ml of NaOH reacted

C= n/V = 1.65 × 10^-3 moles × 1000/12.44

C= 0.13 M

At the equivalence point, the KHP solution turned light pink.

4 0

3 years ago

at a particulat temperature, a smaple of pure water has a Kw of 7.7x10-14. what is the hydroxide concentration of this sample

ycow [4]

Answer: The hydroxide concentration of this sample is $3.85\times 10^{-7}$

Explanation:

When an expression is formed by taking the product of concentration of ions raised to the power of their stoichiometric coefficients in the solution of a salt is known as ionic product.

The ionic product for water is written as:

$K_w=[H^+]\times [OH^-]$

$7.7\times 10^{-14}=[H^+]\times [OH^-]$

As $[H^+]=[OH^-]$

$2[OH^-]=7.7\times 10^{-14}$

$[OH^-]=3.85\times 10^{-7}$

Thus hydroxide concentration of this sample is $3.85\times 10^{-7}$

4 0

2 years ago